Triple Integral over Region: Finding Volume with f(x,y,z) = z

[piepowah]

Homework Statement

Let W be the region bounded by y + z = 2, 2x = y, x = 0, and z = 0. Express and evaluate the triple integral of f (x, y, z) = z by projecting W onto the: (a) xy-plane (b) yz-plane (c) xz-plane.

Homework Equations

The function f (x, y, z) = z and the boundary W: {y + z = 2, 2x = y, x = 0, and z = 0}

The Attempt at a Solution

∫∫∫ z dzdydx; {w: 0 < x < 1, 0 < y < 2x, 0 < z < 2 - y}
w
= ∫∫1/2z^2 dydx z from 0 to 2 - y
= 1/2∫∫ (4 - 4y + y^2)dydx
= ∫(2y - y^2 +1/6y^3) dy y from 0 to 2x
= 2x^2 - 4/3x^3 + 1/3x^4 x from 0 to 1
= 2 - 4/3 + 1/3 = 1

b. yz-plane

∫∫∫ z dxdzdy; {w: 0 < y < 2, 0 < z < 2 - y, 0 < x < y/2}
w
= ∫∫xz/2 dzdy, x from 0 to y/2
= 1/4∫ yz^2 dy, z from 0 to 2 - y
= 1/2y^2 - 1/3y^3 + 1/16y^4, y from 0 to 2
= 2 - 8/3 + 1 = 1/3

c. xz-plane

∫∫∫z dydzdx; {w: 0 < x < 1, 0 < z < 2 - 2x, 2x < y < 2 - z}
w
= ∫∫yz dzdx, y from 2x to 2 - z
= ∫∫(2z - z^2 - 2xz) dzdx
= ∫(z^2 - 1/3z^3 - xy^2) dx, z from 0 to 2 - 2x
= ∫(4 - 8x + 4x^2 - 1/3 (2 - 2x)^3 - x(4 - 8x +4x^2)dx
= (4x - 4x^2 + 4/3x^3 - 2x^2 + 8/3x^3 - x^4), x from 0 to 1
= 4/3 - 2 + 8 /3 - 1 + 16/24
= 5/3

I think that the result is a four-dimensional value, finding the volume over the function
f ( x, y, z) = z. However, the results should come out to be the same value, but in these three cases I don't. I visualize the picture to be a pyramid with a triangular base. I have a feeling I messed up the limits of integration and have tried rewriting the limits of integration many times and have come up with values that don't match up. Can somebody tell me the correct limits of integration for the triple integral and why the ones that I have are incorrect?

Thanks.

 

[haruspex]

Not sure I understand the concept of evaluating a triple integral by projecting the region onto a plane.
As soon as you project it you lose information. E.g. onto the xy plane you lose that y+z=2 is a boundary. Might it mean by taking slices parallel to the given planes?

 

[piepowah]

Yeah, I think by projection the question means slices parallel to the given plane. You can call the projection the domain/boundary of the plane, but it extends with a "height."

 

[haruspex]

Yeah, I think by projection the question means slices parallel to the given plane.

OK, so for the first case, fix z and express the ranges for x and y in terms of that.

 

[piepowah]

By fixing z, do you mean making it constant, from 0 to 2, for the innermost integral?

 

[haruspex]

Yes.

 

[piepowah]

For the first part, the projection is on the xy-plane, so shouldn't dz be first in the order of integration since z depends on both x and y?

Sorry for the late response, it was 1 in the morning where I live so I went to sleep.

 

[LCKurtz]

The way you have written the limits is confusing. Much better to use latex and put the limits on the integrals. That being said, apparently when you write, for example, {w: 0 < x < 1, 0 < y < 2x, 0 < z < 2 - y} you mean for the limits to be arranged like this:$$
\int_{0}^1 \int_0^{2x} \int_0^{2-y} z \, dzdydx$$I don't know what you mean by holding $z$ constant in that first integral though. It varies.

On this first integral, check those middle limits. The other limits look OK so you likely need to check your arithmetic.

[Edit: Added] Once you fix that you should get 1/3 which agrees with your second one. Check your arithmetic on the third one.

 

[haruspex]

For the first part, the projection is on the xy-plane, so shouldn't dz be first in the order of integration since z depends on both x and y?

I read it the other way around. A slice parallel to the xy plane gives you a lamina with a boundary that depends on z. Finding its area means integrating wrt x and y first (whatever order). Integrating wrt z would be the final step.

 

[piepowah]

I finally got the values, 1/3, to match up! Thanks haruspex and LCKurtz for both of your help ^^. I appreciate it.